Partition List

Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.

You should preserve the original relative order of the nodes in each of the two partitions.

For example,
Given 3->2->5->2 and x = 3,
return 1->2->2->4->3->5.

 

思路:

1. 空间复杂度为 o(n) 解法. 创建两个链表, 分别记录大于 x 和小于 x 的节点, 最后合并

2. o(1) 的空间复杂度解法. 四个指针, 分别指向小于 x 部分链表的头, 尾, 指向大于 x 部分链表的头, 尾

 为了简单，我这里使用1的思路。

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    # @param {ListNode} head
    # @param {integer} x
    # @return {ListNode}
    def partition(self, head, x):
       head1=ListNode(0)
       head2=ListNode(0)
       h1flag=head1
       h2flag=head2
       while head!=None:
           if head.val<x:
               h1flag.next=head
               h1flag=h1flag.next
               head=head.next
           else:
               h2flag.next=head
               h2flag=h2flag.next
               head=head.next
       h2flag.next=None           
       h1flag.next=head2.next
       return head1.next